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Competency in Focus: Calculating the limit of the function  This problem from I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016 is based on simple manipulations and limit of a function .

Next understand the problem

Let $f:R \to R$ be a non-zero function such that $\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ exists for all y >0. Let $g(y)= \lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ .If g(1)=1 ,then for all y>0 ?
Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016. Obective Problem no. 21.
Key Competency
Limit of a function
7 out of 10
Suggested Book
Elementary Number Theory by David M. Burton

Do you really need a hint? Try it first!

So , what we have to find in general the value of  g(y) in terms of y  given that $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ provided $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ exists .

See what can we do is that g(y) can be written as follows , $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}= (y^3)\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3y^3}$

Again from the given condition we have $g(1)=\lim_{x \rightarrow \infty} f(x) /x^3=\lim_{xy \rightarrow \infty} \frac{f(xy)}{x^3y^3}=1$.
Therefore , $g(y)=(y^3)\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3y^3} =y^3$ by previous argument .

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